Philosophy of Biology

Formalisations of Evolutionary Biology 513

the sample size times 2 (there are twice as many alleles as individuals) yields the
allelic frequency of this sample. Hence:

The frequency of the + allele = 0.871

The frequency of the ∆32 allele = 0.129

What genotype numbers does the hardy-Weinberg equilibrium yield given these
allelic frequencies?

(p^2 )++:(2pq) + ∆32 : (q^2 ) ∆32∆32

Yields (0. 8712 ) + + : (2(0. 871 X 0 .129)) + ∆32 : (0. 1292 )∆32∆32

=0.758641 + + : 0.224718 + ∆32 : 0.016641∆32∆32

Hence, in a population of 294 individuals, the Hardy-Weinberg equilibrium yields:

++ : 222.9 + ∆32 : 66.2 ∆32∆32 : 4. 9

As we would expect these add up to 294. A comparison of the values expected
based on the Hardy-Weinberg equilibrium and those observed yields:

H−D expected : ++ : 222.9 + ∆32 : 66.2 ∆32∆32 : 4. 9

Observed : ++ : 224 + ∆32 : 64 ∆32∆32 : 6

Now we can ask, how good is the fit between the H-D expected values based on
the specified allelic frequencies and the observed values?
TheX^2 -test is:

X^2 =

∑

(observed quantity – expected quantity)^2 /(expected quantity)

There are three genotypes, hence:

X^2 = ((224− 222 .9)^2 / 222 .9) + ((64− 66 .2)^2 / 66 .2) + ((6− 4 .9)^2 / 4 .9)

=(1. 21 / 222 .9) + (4. 84 / 66 .2) + (1. 21 / 4 .9)

=0.00543 + 0.0731 + 0. 2469

=0. 3254

To use this result to assess goodness of fit, it is necessary to determine the degrees
of freedom for the test.

Degrees of Freedom (df) = (classes of data - 1) – the number of pa-

rameters estimated.

Since there are three genotypes, the classes of data is 3. Sincep+q= 1(hence,q
is a function ofp; they are not independent parameters), there is only 1 parameter
being estimated. Hence, the degrees of freedom for this test is:

(3−1)−1=1